University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 53


$\frac{2xe^{x^2}}{\sqrt {e^{x^2}}}$

Work Step by Step

$y=\int_{0}^{e^{x^2}} ({\frac{1}{\sqrt t}dt}) =\int_{0}^{e^{x^2}} (t^{-1/2}dt) =[2\sqrt t]_{0}^{e^{x^2}}=[2\sqrt {e^{x^2}}]$ Let's differentiate this! $\frac{dy}{dx}=2\cdot\frac{1}{2}({e^{x^2}})^{-1/2}\cdot(2xe^{x^2})=\frac{2xe^{x^2}}{\sqrt {e^{x^2}}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.